2) The curve with the equation )
has a gradient function with the rule =3x^2+k)
, where k is a constant, and has a turning point with the coordinates (-2,6). Find:
the value of k
the rule f(x)
We are told that we have a turning point at (-2,6). One of the definitions of a turning point is that the gradient is 0 i.e
 = 0)
So
 = 0 )
^{2} + k)
 + k )
For the second question, remember this rule
 dx = f(x) )
Therefore:
 = \frac{3x^{3}}{3} - 12x + c )
 = x^{3} - 12x +c )
To find our arbitrary constant (c), we can substitute our co-ordinates (-2,6), because we know that for this particular curve, x = -2, when y=6
=6 )
^{3} - 12(-2) + c )
Therefore, our rule for f(x) is
 = x^{3} - 12x - 10 )
3)Find the area of the region bounded by the graph of y=sin(x), the lines with equations x=pi/6 and x=pi/3 and the x axis.
Remember the formula behind the curve
dx)
where a and b represents the bounded area.
First consider the curve itself (this is where Mao's tablet could come in handy). If there is any area in which the integral would be negative i.e. below x-axis, then you need to separate or cut the area under the curve and work each area separately (including finding out where the area is bounded. When we look at the graph, this does not occur and so we don't need to worry about this.
Our area is bound by the values
and
. We let the lower value be a and the higher value be the b.
So we get this
})dx)
And so you would solve it as if it was a definite integral.
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